Đặt A=\(\dfrac{6}{x+1}.\dfrac{x-1}{3}\)
=\(\dfrac{2}{x+1}.\dfrac{x-1}{1}\)
=\(\dfrac{2.\left(x-1\right)}{x+1}\)
=\(\dfrac{2.x-2}{x+1}\)
=\(\dfrac{2.\left(x+2\right)-4}{x+1}\)
=\(2-\dfrac{4}{x+1}\)
Để A nhận giá trị nguyên thì x + 1 là Ư(4) ={+1;-1;+2;-2;+4;-4}
Suy ra x ∈ {0;-2;1;-3;3;-5}
Đặt \(A=\dfrac{6}{x+1}\times\dfrac{x-1}{3}=\dfrac{2x-2}{x+1}=2-\dfrac{4}{x+1}\)
Để A nguyên \(\Rightarrow\dfrac{4}{x+1}\) nguyên \(\Rightarrow x+1\) là \(Ư\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)
\(x+1=-4\Rightarrow x=-5\)
\(x+1=-2\Rightarrow x=-3\)
\(x+1=-1\Rightarrow x=-2\)
\(x+1=1\Rightarrow x=0\)
\(x+1=2\Rightarrow x=1\)
\(x+1=4\Rightarrow x=3\)
\(\dfrac{6}{x+1}\) và \(\dfrac{x-1}{3}\) là 1 số nguyên khi\(\left\{{}\begin{matrix}6⋮x+1\\x-1⋮3\end{matrix}\right.\)
ta có 6 ⋮ x+1 ⇒ x+1 ∈ Ư(6)
⇒ x+1 ∈ {1;2;3;6}
⇒ x ∈ {0;1;2;5} ⇒ x-1 ∈ {-1;0;1;3}
mà x-1 ⋮ 3 ⇒ x-1 ∈ {0;3}
⇒ x ∈ {1;4}
